Just around every text has this instance of shooting a bullet horizontally and dropping a cartridge from the same height. The idea is the they need to hit the ground in ~ the same time. No one yet the MythBusters can actually present this demo through a actual gun.

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If friend didn't record the recent MythBusters (yeah! new episodes), they go something directly from the physics textbooks. Just around every text has this instance of shoot a bullet horizontally and also dropping a bullet from the very same height. The idea is that they should hit the ground at the same time. No one yet the MythBusters could actually display this demo with a actual gun.


I am going to carry out some calculations, but I want to first write about the physics that accompanies this idea (and you deserve to actually execute it your me without the gun). What physics principle does this demo show? Well, it mirrors two things. First, it shows that the horizontal and also vertical movement in projectile activity are independent. The is, what happens in the y-direction, stays in the y-direction. Really, you can treat projectile movement as 2 separate problems that just take place to take the exact same amount the time. The 2nd thing it reflects is that the velocity in one direction does not impact the velocity in the various other direction (which really is the very same as the first thing). Allow me attract two diagrams, one because that a reduce ball and also one for sphere shot horizontally.

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There space some really essential things to keep in mind in this diagram. First, the pressure is only in the y-direction. This means that the red ball's x-motion does not change and no does the blue ball's. Second, the acceleration in the y-direction for both balls will be the same. Also if the balls had various masses, this would be true due to the fact that a better mass would have actually a greater gravitational pressure from the Earth.


Since the x- and y-motions are independent, and since the balls have actually the same y-motion they would hit the ground in ~ the exact same time.


But what about air resistance?

Yes, this complicates things somewhat. First, ns will usage a usual model because that air resistance that counts on part stuff and the speed of the object. Right here is the model:


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This is the size of the wait resistance. The direction is in the opposite direction together the way the thing moves. So, below is a bullet shot horizontally native a gun through air resistance.

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Here over there is a difference. Currently there is a component of pressure that depends on velocity in the y-direction. Will this do it fight the ground at a different time 보다 one that is just dropped? also if the one likewise has wait resistance? Let's just do it. The course, ns am not the MythBusters. I don't even own a gun. I do own a computer.


Modeling Bullet through air resistance

It seems favor I have been using my bullet version a many lately. Each time I change it just a little. In this case, ns am walking to start with a .45 caliber bullet like the mythbusters did. They claimed they offered it due to the fact that it was heavy and also slow. The is fine through me. Slow means that I have the right to just usage one worth for the traction coefficient instead of my super-sonic coefficient table. If you desire my python program, i will add it in ~ the end. Yet here is my very first run. I am going come shoot a .45 caliber cartridge aimed horizontally 36 inches above the ground. Just as a comparison, I will certainly plot the trajectory of the .45 and also another object shot through it however without air resistance.

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This offers a range of 102 meters (or feet) i beg your pardon is comparable to what the MythBusters found. Notification that there is no air resistance, the object just goes a couple of (7 ish) meter more. Currently how about a to compare of your vertical motions.


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I zoomed in to simply the end. Here you can see that the bullet v air resistance hit the ground after ~ 0.44 seconds. The cartridge without waiting resistance struggle the ground ~ 0.431 seconds. Ok, now to add air resistance come the dropped bullet and also just drop the (don't shoot it). Also, i will add a plot the a falling cartridge without wait resistance.

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They don't hit in ~ the very same time. Actually, all three objects will certainly hit differently. The no air drag object will certainly hit first, climate the dropped bullet and then the fired bullet. The above graph watch like simply two lines because the dropped bullet and also the no traction bullet are so similar. Permit me zoom in a little bit on that graph.

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What if I increase the air resistance? I might redo this through something like a baseball, however I will just boost the coefficient C. Here is what ns get.

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This is through a ginourmous drag coefficient that C = 20. So, is this right? should the to reduce bullet yes, really hit the ground very first with wait resistance? let me draw two force diagrams for a dropped bullet and a fired bullet simply a quick moment after ~ they room released.

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When initially released, both bullets had the same pressure in the y-direction. The fired cartridge was moving horizontally so that the wait resistance was only in the x-direction. So, they both initially had actually the same vertical acceleration and after a brief time around the very same velocity (I will contact this v2y). But, exactly how do the upright air resistances compare? allow me contact the initial launch rate of the cartridge vbullet. One much more thing, i am walk to write the waiting resistance together Kv2 just due to the fact that it is shorter. The wait resistance ~ above the dropped bullet will be:

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Now let me look in ~ the y-component of waiting resistance pressure on the fired bullet. Let me just say the it yes, really hasn't slowed down lot (doesn't really issue anyway) in the x-direction. The size of the totality air resistance pressure would be:

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I left off the tiny little y-component of velocity due to the fact that it is just too little to matter. However what I desire is the upright component of the wait resistance. To execute this, I require the angle below the horizontal the bullet is traveling. This diagram may be useful:

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So, I can write the following:

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And climate the y-component that the wait resistance would be:

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There you have it. The y-component of air resistance for the fired cartridge still counts on the fired speed of the bullet (since that is proportional to v2). A fired bullet (with air resistance) does not hit the ground in ~ the same time as a dropped bullet. That being said, it to be still an great demo. The .45 caliber to be a good choice since it is slow and also heavy and also air resistance doesn't play that large of a role.